Question: $\dfrac{dy}{dx}=y^4$ and $y(2)=-1$. $y(-1)=$
Explanation: The differential equation is separable. What does it look like after we separate the variables? $\dfrac{dy}{y^4}=dx$ Let's integrate both sides of the equation. $\int\dfrac{dy}{y^4}=\int dx$ What do we get? $-\dfrac1{3y^3}=x+C$ What value of $C$ makes the solution curve pass through the point $(2,-1)$ ? Let's substitute $x=2$ and $y=-1$ into the equation and solve for $C$. $\begin{aligned} -\dfrac1{3(-1)^3}&=2+C\\ \\ \\ \\ \dfrac13&=2+C\\ \\ \\ C&=-\dfrac53 \end{aligned}$ Now use this value of $C$ to find $y$ when $x=-1$. $\begin{aligned} -\dfrac1{3y^3}&=-1-\dfrac53\\ \\ \\ -\dfrac1{3y^3}&=-\dfrac83\\ \\ \\ \dfrac1{y^3}&=8\\ \\ \\ y^3&=\dfrac18\\ \\ \\ y&=\dfrac12 \end{aligned}$